| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The given problem can be solved using the Chinese Remainder Theorem, but it can also be solved by inspection or by using a simple algorithm. Let's denote the smallest positive integer that satisfies the given conditions as \(x\). The given conditions can be written as: 1. \(x \equiv 1 \pmod{2}\) 2. \(x \equiv 2 \pmod{3}\) 3. \(x \equiv 3 \pmod{4}\) 4. \(x \equiv 4 \pmod{5}\) 5. \(x \equiv 5 \pmod{6}\) We can observe that the conditions imply that \(x+1\) is divisible by 2, 3, 4, 5, and 6. The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60, which means that \(x+1\) is a multiple of 60. Thus, we can write \(x = 60k - 1\) for some integer \(k\). We need to find the smallest positive integer \(k\) such that \(x\) is positive. The smallest positive integer \(k\) that makes \(x\) positive is \(k = 1\), because for \(k = 0\), \(x = -1\), which is not positive. Thus, \(x = 60 \times 1 - 1 = 59\). Therefore, the smallest positive integer that satisfies the given conditions is \(\boxed{59}\). |
59 |
| deepseekr1 | The smallest positive integer \( N \) that satisfies the given conditions can be found by observing that each remainder is one less than the divisor. Thus, \( N + 1 \) must be a common multiple of 2, 3, 4, 5, and 6. The least common multiple (LCM) of these numbers is calculated as follows: - Prime factorizations: - \( 2 = 2 \) - \( 3 = 3 \) - \( 4 = 2^2 \) - \( 5 = 5 \) - \( 6 = 2 \times 3 \) - LCM is the product of the highest powers of all primes: - \( \text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 \). Thus, \( N + 1 = 60 \), so \( N = 60 - 1 = 59 \). **Verification**: - \( 59 \div 2 = 29 \) remainder \( 1 \) - \( 59 \div 3 = 19 \) remainder \( 2 \) - \( 59 \div 4 = 14 \) remainder \( 3 \) - \( 59 \div 5 = 11 \) remainder \( 4 \) - \( 59 \div 6 = 9 \) remainder \( 5 \) All conditions are satisfied. Therefore, the smallest positive integer is \(\boxed{59}\). |
59 |
| gpto1 | To solve this problem, we'll find the smallest positive integer \( N \) that satisfies the following congruences: 1. \( N \equiv 1 \mod 2 \) 2. \( N \equiv 2 \mod 3 \) 3. \( N \equiv 3 \mod 4 \) 4. \( N \equiv 4 \mod 5 \) 5. \( N \equiv 5 \mod 6 \) **Observation:** Notice that for each modulus \( k \) from 2 to 6, the remainder when \( N \) is divided by \( k \) is \( k - 1 \). This means: \( N \equiv -1 \mod k \) Because \( k - 1 \) is congruent to \( -1 \) modulo \( k \). **Find the Least Common Multiple (LCM):** We can find \( N \) by finding the smallest number that is congruent to \( -1 \) modulo the least common multiple of the moduli (2, 3, 4, 5, 6). First, calculate the LCM of 2, 3, 4, 5, and 6. Prime factorization: - \( 2 = 2 \) - \( 3 = 3 \) - \( 4 = 2^2 \) - \( 5 = 5 \) - \( 6 = 2 \times 3 \) The LCM is the product of the highest powers of all primes involved: - Highest power of 2: \( 2^2 \) (from 4) - Highest power of 3: \( 3^1 \) (from 3) - Highest power of 5: \( 5^1 \) (from 5) So, \( \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \). **Find \( N \):** Since \( N \equiv -1 \mod 60 \), we have: \( N \equiv 59 \mod 60 \) (since \( -1 \mod 60 = 59 \)) Therefore, the smallest positive integer \( N \) is 59. **Verification:** 1. \( 59 \div 2 = 29 \text{ remainder } 1 \) 2. \( 59 \div 3 = 19 \text{ remainder } 2 \) 3. \( 59 \div 4 = 14 \text{ remainder } 3 \) 4. \( 59 \div 5 = 11 \text{ remainder } 4 \) 5. \( 59 \div 6 = 9 \text{ remainder } 5 \) All conditions are satisfied. **Answer:** 59 |
59 |
| GPT-4o | To solve this problem, we need to find the smallest positive integer \( x \) that satisfies the following set of congruences: \[ x \equiv 1 \pmod{2} \] \[ x \equiv 2 \pmod{3} \] \[ x \equiv 3 \pmod{4} \] \[ x \equiv 4 \pmod{5} \] \[ x \equiv 5 \pmod{6} \] Observing the pattern in these congruences, we can rewrite them as: \[ x \equiv -1 \pmod{2} \] \[ x \equiv -1 \pmod{3} \] \[ x \equiv -1 \pmod{4} \] \[ x \equiv -1 \pmod{5} \] \[ x \equiv -1 \pmod{6} \] This means that \( x + 1 \) must be divisible by 2, 3, 4, 5, and 6. Therefore, \( x + 1 \) is a common multiple of these numbers. We need to find the least common multiple (LCM) of these numbers. The prime factorizations of these numbers are: \[ 2 = 2 \] \[ 3 = 3 \] \[ 4 = 2^2 \] \[ 5 = 5 \] \[ 6 = 2 \cdot 3 \] To find the LCM, we take the highest powers of all prime factors involved: \[ \text{LCM}(2, 3, 4, 5, 6) = 2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60 \] Thus, \( x + 1 = 60k \) for some integer \( k \). To find the smallest positive \( x \), we set \( k = 1 \): \[ x + 1 = 60 \cdot 1 = 60 \] \[ x = 60 - 1 = 59 \] Therefore, the smallest positive integer that satisfies all the given conditions is: \[ \boxed{59} \] |
59 |
| llama3.1-70B-Instruct | The smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, a remainder of 4 when divided by 5, and a remainder of 5 when divided by 6 is 59. To find this solution, we can use the Chinese Remainder Theorem (CRT), which states that if we have a system of congruences: x ≡ a1 (mod n1) x ≡ a2 (mod n2) ... x ≡ ak (mod nk) where ni are pairwise coprime (i.e., gcd(ni, nj) = 1 for i ≠ j), then there exists a unique solution modulo N = n1n2...nk. In this case, we have: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) x ≡ 5 (mod 6) We can rewrite these congruences as: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) x ≡ 5 (mod 6) Using the CRT, we can find the solution modulo N = 2 × 3 × 4 × 5 × 6 = 720. The solution is x ≡ 59 (mod 720), which means that the smallest positive integer that satisfies all the given conditions is 59. |
59 |
| llama3.1-405B-Instruct | To find the smallest positive integer that satisfies all the given conditions, we can use the Chinese Remainder Theorem (CRT). First, let's list the conditions: 1. Leaves a remainder of 1 when divided by 2. 2. Leaves a remainder of 2 when divided by 3. 3. Leaves a remainder of 3 when divided by 4. 4. Leaves a remainder of 4 when divided by 5. 5. Leaves a remainder of 5 when divided by 6. We can rewrite these conditions as: 1. x ≡ 1 (mod 2) 2. x ≡ 2 (mod 3) 3. x ≡ 3 (mod 4) 4. x ≡ 4 (mod 5) 5. x ≡ 5 (mod 6) Now, we can apply the CRT to find the smallest positive integer that satisfies all these congruences. The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60. So, we can write: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) x ≡ 5 (mod 6) Using the CRT, we can find the solution modulo 60: x ≡ 59 (mod 60) Therefore, the smallest positive integer that satisfies all the given conditions is: x = 59 So, the solution is 59. |
59 |
| DeepSeek-Coder-V2 | To solve this problem, we need to find the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \\ x \equiv 5 \pmod{6} \end{cases} \] First, observe that each congruence can be rewritten in a form that suggests a pattern: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \\ x \equiv 5 \pmod{6} \end{cases} \] Notice that \( x \equiv k \pmod{k+1} \) for \( k = 1, 2, 3, 4, 5 \). This implies that \( x \) is always 1 less than a multiple of \( k+1 \). Therefore, we can rewrite the system as: \[ \begin{cases} x + 1 \equiv 0 \pmod{2} \\ x + 1 \equiv 0 \pmod{3} \\ x + 1 \equiv 0 \pmod{4} \\ x + 1 \equiv 0 \pmod{5} \\ x + 1 \equiv 0 \pmod{6} \end{cases} \] This means that \( x + 1 \) must be a common multiple of 2, 3, 4, 5, and 6. To find the smallest such \( x + 1 \), we need to determine the least common multiple (LCM) of these numbers. First, we find the prime factorizations: - \( 2 = 2 \) - \( 3 = 3 \) - \( 4 = 2^2 \) - \( 5 = 5 \) - \( 6 = 2 \times 3 \) The LCM is found by taking the highest power of each prime that appears in these factorizations: - The highest power of 2 is \( 2^2 \) (from 4) - The highest power of 3 is \( 3 \) (from 3 and 6) - The highest power of 5 is \( 5 \) (from 5) Thus, the LCM is: \[ \text{LCM}(2, 3, 4, 5, 6) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 \] So, \( x + 1 = 60 \), which means \( x = 60 - 1 = 59 \). Therefore, the smallest positive integer that satisfies all the given conditions is: \[ \boxed{59} \] |
59 |
| Qwen2-72B-Instruct | The given problem can be solved using the Chinese Remainder Theorem, but it can also be solved by inspection or by using a simple algorithm. Let's denote the smallest positive integer that satisfies the given conditions as \(x\). The given conditions can be written as: 1. \(x \equiv 1 \pmod{2}\) 2. \(x \equiv 2 \pmod{3}\) 3. \(x \equiv 3 \pmod{4}\) 4. \(x \equiv 4 \pmod{5}\) 5. \(x \equiv 5 \pmod{6}\) We can observe that the conditions imply that \(x+1\) is divisible by 2, 3, 4, 5, and 6. The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60, which means that \(x+1\) is a multiple of 60. Thus, we can write \(x = 60k - 1\) for some integer \(k\). We need to find the smallest positive integer \(k\) such that \(x\) is positive. The smallest positive integer \(k\) that makes \(x\) positive is \(k = 1\), because for \(k = 0\), \(x = -1\), which is not positive. Thus, \(x = 60 \times 1 - 1 = 59\). Therefore, the smallest positive integer that satisfies the given conditions is \(\boxed{59}\). |
59 |
| baichuan4 | To find the smallest positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem (CRT). The theorem states that if one wants to solve a system of congruences with pairwise coprime moduli, then there is a unique solution modulo the product of the moduli. Given the remainders and divisors, we can set up the following system of congruences: 1. x ≡ 1 (mod 2) 2. x ≡ 2 (mod 3) 3. x ≡ 3 (mod 4) 4. x ≡ 4 (mod 5) 5. x ≡ 5 (mod 6) The moduli (2, 3, 4, 5, 6) are not pairwise coprime, so we cannot directly apply the CRT. However, we can solve this problem manually by considering each congruence in turn. First, let's find the least common multiple (LCM) of all the divisors (2, 3, 4, 5, 6), which is 60. This means that any number that satisfies the conditions will leave the same remainders when divided by these numbers, and the solution will repeat every 60 numbers. Now, we need to find the solution that satisfies all the congruences. We can start by checking numbers from 1 to 60 to see if they satisfy all the conditions. By inspection, we can see that the number 29 satisfies all the conditions: 1. 29 ÷ 2 has a remainder of 1. 2. 29 ÷ 3 has a remainder of 2. 3. 29 ÷ 4 has a remainder of 3. 4. 29 ÷ 5 has a remainder of 4. 5. 29 ÷ 6 has a remainder of 5. Therefore, the smallest positive integer that satisfies the given conditions is 29. |
59 |
| doubao | One way to approach this problem is to find the least common multiple (LCM) of the divisors 2, 3, 4, 5, and 6, and then subtract 1. The LCM of 2, 3, 4, 5, and 6 is 60. So the number we're looking for is 60 - 1 = 59. 59 leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, a remainder of 4 when divided by 5, and a remainder of 5 when divided by 6. So the answer is 59. |
59 |
| minimax | To solve this problem, we can use the Chinese Remainder Theorem (CRT), which provides a way to solve systems of simultaneous congruences with pairwise coprime moduli. However, in this case, the moduli are not pairwise coprime (for example, 4 and 6 are not coprime), so we cannot directly apply the CRT. Instead, we can look for a pattern or use a brute force approach to find the smallest positive integer that satisfies all the given conditions. Let's denote the integer we are looking for as N. The conditions can be written as: 1. N ≡ 1 (mod 2) 2. N ≡ 2 (mod 3) 3. N ≡ 3 (mod 4) 4. N ≡ 4 (mod 5) 5. N ≡ 5 (mod 6) We can start by listing the first few numbers that satisfy each condition and then look for a common number that satisfies all conditions: 1. N ≡ 1 (mod 2) means N is odd: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... 2. N ≡ 2 (mod 3) means N is 2 more than a multiple of 3: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, ... 3. N ≡ 3 (mod 4) means N is 3 more than a multiple of 4: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ... 4. N ≡ 4 (mod 5) means N is 4 more than a multiple of 5: 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, ... 5. N ≡ 5 (mod 6) means N is 5 more than a multiple of 6: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, ... Now, we look for a number that appears in all these lists. We can see that the number 11 appears in the lists for conditions 2, 3, and 5. However, it does not satisfy condition 4 (11 ≡ 1 (mod 5)). We need to find a number that satisfies all five conditions. By examining the lists, we can see that the number 59 appears in the lists for conditions 3, 4, and 5. Let's check if it satisfies condition 1 and 2: 59 ≡ 1 (mod 2) [satisfies] 59 ≡ 2 (mod 3) [satisfies] Since 59 satisfies all five conditions, it is the smallest positive integer that leaves the specified remainders when divided by 2, 3, 4, 5, and 6. Therefore, the solution to the problem is N = 59. |
59 |
| kimi | This problem is a classic example of a Chinese Remainder Theorem (CRT) problem. The CRT states that if you have a system of simultaneous congruences with coprime moduli, there exists a unique solution modulo the product of the moduli. Here's how you can solve it: 1. You have the following system of congruences: \[ \begin{align*} x &\equiv 1 \pmod{2} \\ x &\equiv 2 \pmod{3} \\ x &\equiv 3 \pmod{4} \\ x &\equiv 4 \pmod{5} \\ x &\equiv 5 \pmod{6} \end{align*} \] 2. Notice that for each modulus, the remainder is one less than the modulus itself. This means that if you add 1 to each remainder, you get a number that is divisible by each modulus: \[ \begin{align*} x + 1 &\equiv 0 \pmod{2} \\ x + 1 &\equiv 0 \pmod{3} \\ x + 1 &\equiv 0 \pmod{4} \\ x + 1 &\equiv 0 \pmod{5} \\ x + 1 &\equiv 0 \pmod{6} \end{align*} \] 3. Now, you are looking for the smallest positive integer \( x \) such that \( x + 1 \) is divisible by the least common multiple (LCM) of the moduli (2, 3, 4, 5, 6). 4. Calculate the LCM of the moduli: - LCM(2, 3) = 6 - LCM(6, 4) = 12 (since 4 is \( 2^2 \) and 6 is \( 2 \times 3 \), the LCM is \( 2^2 \times 3 = 12 \)) - LCM(12, 5) = 60 (since 5 is prime, the LCM is \( 12 \times 5 = 60 \)) 5. The smallest positive integer \( x \) such that \( x + 1 \) is divisible by 60 is \( x = 60 - 1 = 59 \). So, the solution to the problem is \( x = 59 \). |
59 |
| wenxinyiyan | To find the smallest positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem, which states that if we have a system of congruences with coprime moduli, then there exists a unique solution modulo the product of these moduli. In this case, our system of congruences is: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) x ≡ 5 (mod 6) However, we cannot directly apply the Chinese Remainder Theorem because the moduli are not all coprime (for example, 2 and 4 are not coprime, and 3 and 6 are not coprime). To solve this, we can break down the congruences with non-coprime moduli into smaller congruences with coprime moduli. The congruence x ≡ 3 (mod 4) implies that x ≡ 1 (mod 2), so we can ignore the first congruence because it is redundant. The congruence x ≡ 5 (mod 6) can be broken down into two congruences: x ≡ 2 (mod 3) and x ≡ 1 (mod 2). Again, the second congruence is redundant because it is implied by the first congruence and the congruence x ≡ 3 (mod 4). So, we are left with the following system of congruences: x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) Now, we can apply the Chinese Remainder Theorem because the moduli 3, 4, and 5 are coprime. To do this, we find the product of the moduli: 3 * 4 * 5 = 60 Next, we find the partial products: 60 / 3 = 20 60 / 4 = 15 60 / 5 = 12 Now, we find the inverses of these partial products modulo the corresponding moduli: 20^-1 ≡ 2 (mod 3) because 20 * 2 = 40 ≡ 1 (mod 3) 15^-1 ≡ 3 (mod 4) because 15 * 3 = 45 ≡ 1 (mod 4) 12^-1 ≡ 3 (mod 5) because 12 * 3 = 36 ≡ 1 (mod 5) Finally, we compute the solution using the Chinese Remainder Theorem: x ≡ 2 * 20 * 2 + 3 * 15 * 3 + 4 * 12 * 3 (mod 60) x ≡ 80 + 135 + 144 (mod 60) x ≡ 359 (mod 60) x ≡ 59 (mod 60) So, the smallest positive integer that satisfies the given conditions is 59. |
59 |
| zhipu | To solve this problem, we are looking for the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{align*} x &\equiv 1 \pmod{2}, \\ x &\equiv 2 \pmod{3}, \\ x &\equiv 3 \pmod{4}, \\ x &\equiv 4 \pmod{5}, \\ x &\equiv 5 \pmod{6}. \end{align*} \] These congruences can be rewritten in a more revealing way: \[ \begin{align*} x &\equiv 1 \pmod{2}, \\ x &\equiv 2 \pmod{3}, \\ x &\equiv -1 \pmod{4}, \\ x &\equiv -1 \pmod{5}, \\ x &\equiv -1 \pmod{6}. \end{align*} \] Notice that the last three congruences all imply that \( x \equiv -1 \) modulo 4, 5, and 6. Since 4, 5, and 6 are pairwise coprime (they have no common factors other than 1), we can combine these congruences into a single congruence using the Chinese Remainder Theorem: \[ x \equiv -1 \pmod{\text{lcm}(4, 5, 6)}, \] where \(\text{lcm}\) denotes the least common multiple. Calculating the least common multiple of 4, 5, and 6: \[ \text{lcm}(4, 5, 6) = \text{lcm}(\text{lcm}(4, 5), 6) = \text{lcm}(20, 6) = 60. \] Thus, we have: \[ x \equiv -1 \pmod{60}. \] This means that \( x = 60k - 1 \) for some integer \( k \). Now, we need to find the smallest \( k \) such that \( x \) also satisfies the first two congruences: \[ \begin{align*} 60k - 1 &\equiv 1 \pmod{2}, \\ 60k - 1 &\equiv 2 \pmod{3}. \end{align*} \] For the first congruence, since \( 60k \) is always even, \( 60k - 1 \) is always odd, which means it is always congruent to 1 modulo 2. So, the first congruence is automatically satisfied. For the second congruence: \[ 60k - 1 \equiv 2 \pmod{3} \implies 60k \equiv 3 \pmod{3} \implies 0 \equiv 3 \pmod{3}, \] which is always true because 60 is a multiple of 3. Therefore, the smallest \( x \) that satisfies all the given conditions is when \( k = 1 \): \[ x = 60 \cdot 1 - 1 = 59. \] So, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, a remainder of 4 when divided by 5, and a remainder of 5 when divided by 6 is \( \boxed{59} \). |
59 |
| gpto1 | To solve this problem, we'll find the smallest positive integer \( N \) that satisfies the following congruences: 1. \( N \equiv 1 \mod 2 \) 2. \( N \equiv 2 \mod 3 \) 3. \( N \equiv 3 \mod 4 \) 4. \( N \equiv 4 \mod 5 \) 5. \( N \equiv 5 \mod 6 \) **Observation:** Notice that for each modulus \( k \) from 2 to 6, the remainder when \( N \) is divided by \( k \) is \( k - 1 \). This means: \( N \equiv -1 \mod k \) Because \( k - 1 \) is congruent to \( -1 \) modulo \( k \). **Find the Least Common Multiple (LCM):** We can find \( N \) by finding the smallest number that is congruent to \( -1 \) modulo the least common multiple of the moduli (2, 3, 4, 5, 6). First, calculate the LCM of 2, 3, 4, 5, and 6. Prime factorization: - \( 2 = 2 \) - \( 3 = 3 \) - \( 4 = 2^2 \) - \( 5 = 5 \) - \( 6 = 2 \times 3 \) The LCM is the product of the highest powers of all primes involved: - Highest power of 2: \( 2^2 \) (from 4) - Highest power of 3: \( 3^1 \) (from 3) - Highest power of 5: \( 5^1 \) (from 5) So, \( \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \). **Find \( N \):** Since \( N \equiv -1 \mod 60 \), we have: \( N \equiv 59 \mod 60 \) (since \( -1 \mod 60 = 59 \)) Therefore, the smallest positive integer \( N \) is 59. **Verification:** 1. \( 59 \div 2 = 29 \text{ remainder } 1 \) 2. \( 59 \div 3 = 19 \text{ remainder } 2 \) 3. \( 59 \div 4 = 14 \text{ remainder } 3 \) 4. \( 59 \div 5 = 11 \text{ remainder } 4 \) 5. \( 59 \div 6 = 9 \text{ remainder } 5 \) All conditions are satisfied. **Answer:** 59 |
59 |
| deepseekv3 | To solve the problem, we need to find the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \\ x \equiv 5 \pmod{6} \end{cases} \] ### Step 1: Observe the Pattern Notice that each congruence can be rewritten as: \[ x \equiv -1 \pmod{2} \\ x \equiv -1 \pmod{3} \\ x \equiv -1 \pmod{4} \\ x \equiv -1 \pmod{5} \\ x \equiv -1 \pmod{6} \] This means that \( x + 1 \) is divisible by 2, 3, 4, 5, and 6. ### Step 2: Find the Least Common Multiple (LCM) The smallest positive integer \( x + 1 \) that is divisible by 2, 3, 4, 5, and 6 is the least common multiple (LCM) of these numbers. \[ \text{LCM}(2, 3, 4, 5, 6) \] First, factorize each number: \[ 2 = 2 \\ 3 = 3 \\ 4 = 2^2 \\ 5 = 5 \\ 6 = 2 \cdot 3 \] The LCM is the product of the highest powers of all prime factors: \[ \text{LCM} = 2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60 \] ### Step 3: Solve for \( x \) Since \( x + 1 = 60 \), we subtract 1 to find \( x \): \[ x = 60 - 1 = 59 \] ### Step 4: Verify the Solution Check that 59 satisfies all the original congruences: \[ 59 \div 2 = 29 \text{ with a remainder of } 1 \\ 59 \div 3 = 19 \text{ with a remainder of } 2 \\ 59 \div 4 = 14 \text{ with a remainder of } 3 \\ 59 \div 5 = 11 \text{ with a remainder of } 4 \\ 59 \div 6 = 9 \text{ with a remainder of } 5 \] All conditions are satisfied. ### Final Answer: The smallest positive integer that satisfies the given conditions is \( \boxed{59} \). |
59 |